TOPIC 3 Motion and gravitational Fields

Transcription

1 TOPIC 3 Motion and gravitational Fields 3.1 Overview Module 5: Advanced Mechanics Motion in gravitational fields Inquiry question: How does the force of gravity determine the motion of planets and satellites? Students: apply qualitatively and quantitatively Newton s Law of Universal Gravitation to: GMm determine the force of gravity between two objects F = ( 2 r ) GM investigate the factors that affect the gravitational field strength g = ( 2 r ) predict the gravitational field strength at any point in a gravitational field, including at the surface of a planet (ACSPH094, ACSPH095, ACSPH097) investigate the orbital motion of planets and artificial satellites when applying the relationships between the following quantities: gravitational force centripetal force centripetal acceleration mass orbital radius orbital velocity orbital period predict quantitatively the orbital properties of planets and satellites in a variety of situations, including near the Earth and geostationary orbits, and relate these to their uses (ACSPH101) investigate the relationship of Kepler s Laws of Planetary Motion to the forces acting on, and the total energy of, planets in circular and non-circular orbits using: (ACSPH101) v 0 = 2πr T r3 T 2 = GM 4π 2 derive quantitatively and apply the concepts of gravitational force and gravitational potential energy in radial gravitational fields to a variety of situations, including but not limited to: 2GM the concept of escape velocity v esc = ( r ) total potential energy of a planet or satellite in its orbit U = GMm ( r ) total energy of a planet or satellite in its orbit E = GM ( 2r ) energy changes that occur when satellites move between orbits (ACSPH096) Kepler s Laws of Planetary Motion (ACSPH101) TOPIC 3 Motion and gravitational Fields 59

2 FIGURE 3.1 Understanding gravitational forces has allowed us to put satellites in orbit around the Earth. 3.2 Explaining the solar system The heliocentric solar system Up until the dawn of the Renaissance, the widely held view of the solar system s mechanics placed the Earth firmly in the centre, with the Sun, Moon and other planets moving in circular orbits around it. This view was overturned by the work of several key figures during what is now considered to be the Golden Age of Science. In 1542, Nicolas Copernicus ( ) published On the Revolution of the Heavenly Orbs, outlining an explanation for the observations of planetary motion with the Sun at the centre. In his explanation, the planets moved in circular orbits about the Sun. Copernicus s model became increasingly preferred over the geocentric model of Ptolemy because it made astronomical and astrological calculations easier. The publication had a significant scientific, social and political impact during the latter part of the 16th century. Galileo Galilei ( ) was a strong advocate for the view that the Copernican model was more than a set of mathematical contrivances, merely to provide a correct basis for calculation and instead represented physical reality. (This had also been Copernicus s view, but he could not express this in print.) Galileo thought that astronomy could now ask questions about the structure, fabric and operation of the heavens, but as with so many of his scientific interests, Galileo did not pursue these questions further. Johannes Kepler ( ) decided his purpose in life was to reveal the fundamental coherence of a planetary system with the Sun as its centre. In he was working as an assistant to Tycho Brahe ( ), a Danish astronomer who had been compiling very precise measurements of the planets positions for over twenty years. Brahe s measurements were so accurate that they are comparable to those achieved with today s more sophisticated technology. Kepler was seeking to find patterns and relationships between motion of the various planets. He used Brahe s data to calculate the positions of the planets as they would be observed by someone outside the 60 Jacaranda Physics 12

3 solar system, rather than from the revolving platform of the Earth. Initially he was looking for circular orbits, but Brahe s precise data did not fit such orbits Kepler s First Law The Law of Ellipses After Kepler used a variety of other geometric shapes to model the orbits of the planets, in 1604 he formulated what is known as Kepler s First Law: Each planet moves, not in a circle, but in an ellipse, with the sun, off centre, at a focus. Ellipses can be thought of as elongated or stretched circles; the degree of stretch is known as eccentricity. The eccentricity of an ellipse is defined as the ratio c, where c is the distance between the two foci of the 2a ellipse and a is the semi-major axis (see Figure 3.2). A circle is an ellipse with an eccentricity of zero. FIGURE 3.2 Apogee furthest point Focus a c The geometry of an ellipse. b Focus Perigee closest point a = semi-major axis b = semi-minor axis c = distance between foci c Eccentricity = 2a FIGURE 3.3 Kepler stated that for a planet s elliptical orbit, the Sun is at one focus and nothing is located at the other focus. Planet Most of the planets in our solar systems have orbits with small eccentricities, as shown in Table 3.1. In contrast to the planets, most comets and some asteroids have very eccentric orbits. Our own Moon has an orbit with an eccentricity of , although this can vary due to a number of influences, not the least of which is the strong gravitational attraction of the Sun. TABLE 3.1 Eccentricities of planetary orbits. Planet Eccentricity Mercury Venus Earth Mars Jupiter Saturn Uranus Try this Neptune An ellipse is like a stretched circle. The shape can be drawn by placing two pins on the page several centimetres apart, with a loose piece of string tied between the pins. If a pencil is placed against the string to keep it tight and then the pencil is moved around the page, the drawn shape is an ellipse with a focus at each of the pins. The closer the two foci, the more like a circle the ellipse becomes. Sun Foci TOPIC 3 Motion and gravitational Fields 61

4 FIGURE 3.4 Drawing an ellipse. Sun FIGURE 3.5 Location of the two equinoxes in the Earth s orbit; note that the elliptical path of the Earth s orbit is greatly exaggerated in this diagram for clarity. September equinox March equinox Kepler s theory of the elliptical nature of planetary orbits was supported by the timing of the Earth s solar equinoxes. The equinoxes are the two days in the year when the Sun is directly above the equator and the durations of night and day are equal. They occur when the line drawn from the Sun to the Earth is at right angles to the Earth s orbit. If the orbit of the Earth around the Sun was a perfect circle, then the time periods elapsing between the equinoxes would be of equal length. This is not the case. In fact, the time from the March equinox until the September equinox is several days longer than the time period from the September equinox to the next March equinox. This difference is consistent with an elliptical orbit where, as the Sun is located at one of two foci, the two points at which the equinoxes occur will not be directly opposite each other Kepler s Second Law the Law of Equal Areas Kepler also looked at the speed of the planets in their orbits. His analysis of the data showed that speeds of the planets FIGURE 3.6 Kepler s Second Law. The regions swept out by the planet during each were not constant. The planets were slower when they were month are equal in area. further away from the Sun and faster when closer. He also found that their angular speed the number of degrees a One month line from the Sun to planet sweeps through every day was not constant. These observations supported his theory as to the elliptical nature of planetary orbits. Further analysis Sun revealed that, while the planets covered different distances in equal time periods, each planet swept out the same area Planet One month per unit time. This is summarised in Kepler s Second Law published in 1609: The linear speed and angular speed of a planet are not constant, but the areal speed of each planet is constant. That is, a line joining the sun to a planet sweeps out equal areas in equal times. Sun 62 Jacaranda Physics 12

5 3.2.4 Kepler s Third Law The Law of Periods Kepler was keen to find a mathematical relationship between the period of a planet s orbit around the Sun and its average radius that gave the same result for each planet. He tried numerous possibilities and eventually, in 1619, he found a relationship that was consistent with the data. This is Kepler s Third Law: For all planets, the cube of the average radius is proportional to the square of the orbital period; that is, R 3 is a constant for all planets going around the sun. T2 Kepler was also able to show that the relationship held for the orbits of the moons of Jupiter. Kepler had constructed as detailed a description of the solar system as was possible without a mechanism to explain the motion of the planets, although he did understand gravity as a reciprocal attraction. Kepler wrote, Gravity is the mutual tendency between bodies towards unity or contact (of which the magnetic force also is), so that the Earth draws a stone much more than the stone draws the Earth... TABLE 3.2 The solar system: some useful data. Body Mass (kg) Radius of body (m) Mean radius of orbit (m) Period of revolution (s) Sun Not applicable Not applicable Earth Moon Mercury Venus Mars Jupiter Saturn Uranus Neptune Pluto* *Pluto is no longer classified as a planet. Scientists have recently hypothesised that a ninth planet may exist, but it has not yet been directly observed. 3.2 SAMPLE PROBLEM 1 Use the information in table 3.2 to determine the average time taken for a satellite to orbit the Earth at an altitude of 500 km from its surface. SOLUTION: As the satellite and the Moon both orbit the Earth, they will have the same ratio for the value of R3 where R is the orbital radius for each object s orbit. Note that this is measured between their centres of mass. R satellite = satellite altitude + radius of the Earth = m m = m R Moon = m T Moon = s T 2 TOPIC 3 Motion and gravitational Fields 63

6 Given that R 3 Moon T 2 Moon = R3 satellite T 2 satellite T satellite = R 3 satellite T2 Moon R 3 Moon ( m) 3 ( s) 2 = ( m) 3 = s The satellite has an orbital period of s (approximately 1 hour and 34 minutes). 3.2 Exercise 1 1 Which of the following statements is true? (a) The orbits of planets around the Sun are in the shape of slightly squashed circles. (b) Planets move faster in their orbits the further away they are from the Sun. (c) The Sun is in the exact geometric centre of our solar system. (d) A planet that is twice as far away from the Sun as the Earth will take twice as long as the Earth to complete its orbit. 2 An AU (Astronomical Unit) is defined as being equal to the average orbital radius of the Earth from the Sun. Given that the Venus year is equal to Earth years, determine its distance from the Sun in AU. 3 (a) Use the information in table 3.2 to determine the value of the ratio R3 for i. Mars, and ii. Jupiter. Give your answers to 2 significant figures. (b) Would you expect this ratio to have the same value for all planets in our solar system? Explain your answer. 4 Ganymede and Io are two of Jupiter s moons. The orbital period of Ganymede around Jupiter is 4.1 times that of Io. If Io has an average orbital radius of m, determine the orbital radius of Ganymede. Watch this elesson: Kepler s Laws Searchlight ID: eles Newton s Law of Universal Gravitation The Apple and the Moon With his three laws and Brahe s observational data, Kepler was able to accurately calculate how the solar system was arranged and how it moved. However, he could not say why it moved this way. This understanding would have to wait another 80 years for the work of Isaac Newton. Isaac Newton is one of the most important people in the history of science, a genius whose influence has been felt in many varied fields within physics. In 1687, he published an enormously important work titled T 2 64 Jacaranda Physics 12

7 Philosophiae Naturalis Principia Mathematica, which contains theories on the motions of objects, as well as a new type of mathematics calculus needed to analyse these motions. In this work, Newton described how his observation of an apple that had been pulled from the tree by the attractive force of the Earth led him to consider whether the same effect (which he described as action at a distance ) would be observed if the apple were falling from a very great height as high as the Moon, say. Newton had previously developed expressions for the inward acceleration of objects that had uniform circular motion. These expressions will be familiar to you from your earlier study of centripetal motion, specifically: a c = v2 r and a c = 4πr T 2 where a c is the centripetal acceleration of an object moving uniformly in a circle with radius r with an orbital speed v, and having an orbital period of T. Newton theorised that the same agency that pulled the apple to the Earth also provided the centripetal acceleration that pulled the Moon into its circular path. Newton compared the acceleration of the Moon and of the apple using the accepted scientific values of the time: Radius of the Earth, R E = 6371 km Earth Moon distance, r = 60 R E Period of the Moon, T = 27.3 days = s a apple = 9.8 m s 2 a Moon = v2 Moon 2πr 2 1 = r ( T ) r = 4π2 r a apple = 4π2 r = 4π2 (60R E ) T 2 T 2 = 4π2 (60 ( m)) ( s) 2 = m s m s 2 = = 3616 a Moon m s 2 T 2 Then, he considered the relative distances of the apple and the Moon from the Earth s centre: d Moon d apple = r R E = 60R E =60 R E Given that 3616 = , it could be seen that d Moon ( d apple ) 2 a apple ( a Moon ) As an acceleration is the result of a force being applied and the acceleration is directly proportional to the size of the force (Second Law of Motion) this implied that 2 d Moon F apple ( d apple ) ( F Moon ) TOPIC 3 Motion and gravitational Fields 65

8 where F apple and F Moon are the attractive gravitational forces that the Earth exerts on the apple and the Moon respectively. In this way, Newton concluded that the size of the gravitational force that the Earth exerts on any object is inversely proportional to that object s distance (r) from the Earth s centre of mass, that is: F g 1 r Newton s Law of Universal Gravitation While other scientists, such as Edmund Halley and Robert Hooke, had theorised that the force that the Sun exerted on a planet or a comet was inversely proportional to the square of the distance between them, Newton was the first to prove it mathematically and to realise that the force involved was gravity the same force that pulled all terrestrial objects downwards. Newton combined his deductions from Kepler s laws with his own laws of motion to develop an expression for a law of universal gravitation that would allow the gravitational attraction between any two objects an apple and a planet, a moon and a planet, a planet and a Sun to be calculated. Using his second law of motion, F net = ma, Newton reasoned that the gravitational force exerted on a planet by the Sun depended on the mass of the planet. By the third law of motion, the force that the Sun exerted on a planet must be equal in magnitude (but opposite in direction) to the force that the planet exerted on the Sun. By the second law of motion, the magnitude of the force that the planet exerted on the Sun depended on the mass of the Sun. Combining these ideas with the inverse square law for distance, Newton derived a relationship for the magnitude of the gravitational force acting between the planet and the Sun: F Sun on planet m Sun m planet r 2 From this, he developed the general equation for the Law of Universal Gravitation: F g = G m 1 m 2 r 2 where G is the universal gravitational constant and m 1 and m 2 are the masses of any two objects. Note that gravitation is always an attractive force. For this reason, you may sometimes see this equation expressed as F g = G m 1 m 2 r 2 (You will see where this negative sign comes from later, in section ) You may also see this equation written as: F g = G M m r 2 where the gravitational force between a very small mass (m) and a much larger mass (M) is being calculated. Regardless of the size of the masses involved, the gravitational force is exerted equally on both masses. The value of G could not be determined at the time because the mass of the Earth was not known. It took another 130 years before Henry Cavendish was able to measure the gravitational attraction between two known masses and calculate the value of G to be N m 2 kg Jacaranda Physics 12

9 The value of G is very small, which indicates that gravitation is quite a weak force. A large quantity of mass is need to produce a gravitational effect that is easily noticeable. 3.3 SAMPLE PROBLEM 1 Given the following data, determine the magnitude of the gravitational attraction between: (a) the Earth and the Moon (b) the Earth and the Sun. mass of the Earth = kg mass of the Moon = kg mass of the Sun = kg Average Earth Moon distance = m Average Earth Sun distance = m SOLUTION: (a) (Note: This distance is referred to as one astronomical unit or 1 AU) F g = Gm Em M r 2 = N m 2 kg 2 ( kg) ( kg) ( m) 2 = N That is, the magnitude of the gravitational force of attraction between the Earth and the Moon is approximately N. (b) F g = Gm E m S r 2 = N m 2 kg 2 ( kg) ( kg) ( m) 2 = N That is, the magnitude of the gravitational force of attraction between the Earth and the Sun is approximately N, or about 180 times greater than the Earth Moon attraction Universal Gravitation and Kepler s Law of Periods Newton s law of universal gravitation placed the physics of planetary orbits on a firm mathematical footing, as well as explaining why the planets and moons move as they do. From this law, Kepler s law of periods can be derived. Kepler had originally stated his Law of Periods in the form R3 = k, but he was not able to determine an T 2 expression for the constant k. When Isaac Newton was devising his Law of Universal Gravitation, he found that he was able to derive such an expression. The derivation begins by equating the gravitation and centripetal forces: F g = F c G M m r 2 = mv2 r TOPIC 3 Motion and gravitational Fields 67

10 where M is the mass of the Sun, m is the mass of the planet, r is the distance between the centres of the planet and the Sun and v is the planet s orbital speed. As v = 2πr T G M m r 2 G m M r 2, where T is the planet s period, = m 2 r (2πr T ) = 4π2 m r T 2 = r3 G M 4π 2 T = k 2 Thus, Kepler s constant k depends only on the mass of the Sun and thus has the same value for all planets orbiting the Sun. Similarly, the Moon and all other satellites orbiting the Earth will have the same value for R3 T. 2 although in this case the value will equal G M Earth. 4π SAMPLE PROBLEM 2 Calculate the value of R3 for the Moon using the data in Sample Problem 1 and then use that value to 2 T calculate the mass of the Earth. SOLUTION: Radius of Moon s orbit, r = m; period, T = s; G = N m 2 kg 2 ; mass of Earth, M Earth =? k = r3 T 2 = ( m) 3 ( s) 2 = m 3 s 2 As k = G m Earth, 4π 2 m Earth = 4π2 k G = 4π2 ( m 3 s 2 ) N m 2 kg 2 = kg 68 Jacaranda Physics 12

11 3.3 Exercise 1 In the following exercises, refer to Table 3.2 for planetary values. 1 Calculate the force due to gravity of the Earth on a 70 kg person standing on the equator. 2 Calculate the force due to gravity by: (a) the Earth on the Moon (b) the Moon on Earth. 3 Determine the gravitational force of attraction between two 250 g apples that are lying on a desk 1.0 m apart. 4 Determine the gravitational force of attraction between Jupiter ( kg) and its moon Ganymede ( kg). The orbital radius of Ganymede is 107 million km. 5 Calculate the gravitational force between Ganymede and a 250 g apple lying on Ganymede s surface. The radius of Ganymede is 2631 km. 6 What is the net gravitational force acting on the Moon during a lunar eclipse? (Assume that the centres of the Sun, Moon and Earth are in a straight line). 7 Two spheres with masses of 8 kg and 6 kg are separated by a distance of 50 cm. At what position would a 2 kg sphere need to be placed between them to experience a net gravitational force of zero? (Ignore the local gravitational field.) 8 Two small space rocks deep in interstellar space are separated by a distance of 1 m. If one of the space rocks has a mass of 200 g while the other has a mass of 150 g, how long will it take for them to be pulled together by gravitational attraction? Ignore the radius of the rocks themselves and assume that they are initially at rest. 3.4 Objects in orbit Determining orbital speed Newton s Law of Universal Gravitation can also be used to calculate the average orbital speed (v o ) of a planet around the Sun: F c = F g m planet v 2 0 = G M Sun m planet r r 2 Dividing both sides by m planet : v 2 0 r = G M Sun r 2 v 2 0 = G M Sun r G M v 0 = Sun r Note that the value of a satellite s orbit depends on: the mass of the planet being orbited the radius of the orbit. For a satellite orbiting a planet, this is equal to the radius of the planet plus the altitude of the orbit. TOPIC 3 Motion and gravitational Fields 69

12 Hence, for the case of a satellite orbiting the Earth, the formula becomes: G M v 0 = E R Earth + altitude where v o = orbital velocity (m s 1 ); M E = mass of the Earth = kg; R E = radius of the Earth = m; altitude = height of orbit above the ground (m). It is clear from this formula that altitude is the only variable that determines the orbital velocity required for a specific orbit. Further, the greater the radius of the orbit, the lower that velocity is. 3.4 SAMPLE PROBLEM 1 Calculate the periods of three different satellites orbiting the Earth at altitudes of (a) 250 km, (b) 400 km and (c) km. SOLUTION: M E = kg; R E = m; G = N m 2 kg 2 (a) At an altitude of 250 km: G M v 0 = E R Earth + altitude = N m 2 kg 2 ( kg) ( m) + ( m) = m s km h 1 (b) At an altitude of 400 km: G M v 0 = E R Earth + altitude = N m 2 kg 2 ( kg) ( m) + ( m) = m s km h 1 (c) At an altitude of km: G M v 0 = E R Earth + altitude = N m 2 kg 2 ( kg) ( m) + ( m) = m s km h 1 70 Jacaranda Physics 12

13 3.4.2 Elliptical orbits The preceding theory assumes that we are dealing with circular orbits but that is usually not the case. As we have seen earlier in this chapter, the most common orbital shape is an ellipse, which can be round or elongated with the degree of elongation described by its eccentricity e: e = c where c is the distance between the two 2a foci of the ellipse and a is the semi-major axis. Most satellites are placed into near-circular orbits, as shown in table 3.3, but there are a few notable exceptions which will be discussed later in this section. Much of the information shown in table 3.3 can be calculated if the semi-major axis a is known, and this can be determined from the apogee and perigee distances. TABLE 3.3 A sample of various Earth satellites. Satellite name GENESAT USA 197 IRIDIUM 95 NOAA 18 RASCOM 1 MOLNIYA 3 53 NAVSTAR 59 OPTUS D2 Purpose Biological research and amateur radio beacon Military eyein-the-sky Satellite phone communication Weather African communications Global Positioning System TV and military communications Australian and NZ television communications Orbit description Period (min) Inclination (degrees) Perigee altitude (km) Low Earth orbit Polar low Earth orbit Velocity at perigee (kmh 1 ) Apogee altitude(km) Velocity at apogee(kmh 1 ) Eccentricity Low Earth orbit Polar low Earth orbit Transfer orbit to geostationary position Elliptical Molniya orbit High altitude GPS orbit Geostationary orbit SKYNET 5B Military communications Geostationary orbit TOPIC 3 Motion and gravitational Fields 71

14 The velocity of a satellite at any point along an elliptical path can be calculated using the following general equation: FIGURE 3.7 The features of the elliptical orbit of a satellite around the Earth. 2 v = G M Earth ( r 1 a) where r is the orbital radius at the point being considered. Earth a The satellite velocities at apogee and perigee in table 3.3 were calculated using this formula (You should confirm that for a circular orbit, a = r Apogee and that this formula simplifies to the orbital velocity equation.) Notice furthest how the velocities of each satellite are least at the apogee and greatest at point the perigee, that is, the satellites move quickly when closest to the Earth a = semi-major axis and slow down as they move further away. This, of course, is just what is b = semi-minor axis c = distance between foci described in Kepler s second law. c Eccentricity = The Molniya orbit was developed specifically for this reason. Devised 2a for Russian communications (as most of Russia lies too far north to be satisfactorily covered by a geostationary satellite), this very eccentric orbit places a high apogee over the desired location. A Molniya satellite will cruise slowly through this apogee before zipping around through the low perigee and returning quickly to the coverage area Types of orbit Focus c b Focus Perigee closest point Spacecraft or satellites placed into orbit will generally be placed into one of two altitudes either a low Earth orbit or a geostationary orbit. A low Earth orbit is generally an orbit higher than approximately 250 km, in order to avoid atmospheric drag, and lower than approximately 1000 km, which is the altitude at which the Van Allen radiation belts start to appear. These belts are regions of high radiation trapped by the Earth s magnetic field and pose significant risk to live space travelers as well as to electronic equipment. The space shuttle utilised a low Earth orbit somewhere between 250 km and 400 km depending upon the mission. At 250 km, an orbiting spacecraft has a velocity of km h 1 and takes just 90 minutes to complete an orbit of the Earth. A geostationary orbit is at an altitude at which the period of the orbit precisely matches that of the Earth. If over the equator, such an orbit would allow a satellite to remain parked over a fixed point on the surface of the Earth throughout the day and night. From the Earth, such a satellite appears to be stationary in the sky, always located in the same direction regardless of the time of day. This is particularly useful for communications satellites because a receiving dish need only point to a fixed spot in the sky in order to remain in contact with the satellite. The altitude of such an orbit can be calculated from Kepler s Law of Periods. Firstly, the period of the orbit must equal the length of one sidereal day; that is, the time it takes the Earth to rotate once on its orbit, relative to the stars. This is 3 minutes and 56 seconds less than a 24-hour solar day, so that T is set to be s. The radius of the orbit then works out to be km, or 6.61 Earth radii. Subtracting the radius of the Earth gives the altitude as approximately km. This places the satellite at the upper limits of the Van Allen radiation belts and near the edge of the magnetosphere, making them useful for scientific purposes as well. Australia has the AUSSAT and OPTUS satellites in geostationary orbits. If a satellite at this height is not positioned over the equator but at some other latitude, it will not remain fixed at one point in the sky. Instead, from the Earth the satellite will appear to trace out a figure of eight path each 24 hours. It still has a period equal to Earth s, however, so this orbit is referred to as geosynchronous. A transfer orbit is a path used to manoeuvre a satellite from one orbit to another. Satellites headed for a geostationary orbit are first placed into a low Earth orbit and then boosted up from there using a transfer orbit, which has a specific orbital energy that lies between that of the lower and higher circular orbits. Orbital 72 Jacaranda Physics 12

15 manoeuvres utilise Keplerian motion, which is not always intuitive. In order to move a satellite into a different orbit, the satellite s energy must be changed; this is achieved by rapidly altering the kinetic energy. Rockets are fired to change the satellite s velocity by a certain amount, referred to as delta-v (Δv), which will increase or decrease the kinetic energy (and therefore the total energy) to alter the orbit as desired. However, as soon as the satellite begins to change altitude, transformations between the potential energy and the kinetic energy occur, so its speed is continually changing. FIGURE 3.8 A Hohmann transfer orbit used to raise a satellite from a low Earth orbit of altitude 400 km up to a geostationary orbit of altitude km. Apogee v = 5820 km h 1 Δv = 5280 km h 1 Δv = 8600 km h 1 Geostationary orbit v = km h 1 Transfer ellipse Low Earth orbit v = km h 1 Earth Perigee v = km h 1 The spacecraft slows when moving out to apogee, though total E does not change. The simplest and most fuel-efficient path is a Hohmann transfer orbit, as shown in figure 3.8. This is a transfer ellipse that touches the lower orbit at its perigee and touches the higher orbit at its apogee. The Hohmann transfer involves two relatively quick (called impulsive ) rocket boosts. In orbital mechanics, the word impulsive describes a quick change in velocity and energy. In order to move to a higher orbit, the first boost increases the satellite s velocity, stretching the circular low Earth orbit out into a transfer ellipse. The perigee is the fastest point on this transfer orbit and, as the satellite moves along the ellipse, it slows again. When it finally reaches the apogee, it will be at the correct altitude for its new orbit, but it will be moving too slowly, the apogee being the slowest point on the ellipse. At this point the rockets are fired again, to increase the velocity to that required for the new higher, stable and circular orbit. TOPIC 3 Motion and gravitational Fields 73

16 In order to move down from a higher to a lower orbit, the process is reversed, requiring two negative delta-v rocket boosts, that is, retro-firing of the rockets. These two boosts will slow the satellite, hence changing its orbit, first from the higher circular orbit into a transfer ellipse that reaches down to lower altitudes, then from the perigee of the transfer ellipse into a circular low Earth orbit. 3.4 Exercise 1 1 Compare, in words only, low Earth orbits and geostationary orbits. 2 Distinguish between a geostationary orbit and a geosynchronous orbit. 3 Calculate the altitude, period and velocity data to complete the following table. TABLE 3.4 Type of orbit Altitude (km) Period (h) Velocity (m s 1 ) Low Earth 360 Geostationary 24 4 (a) Define the orbital velocity of a satellite. (b) Describe its relationship to: i. the gravitational constant G ii. the mass of the planet it is orbiting iii. the mass of the satellite iv. the radius of the orbit v. the altitude of the satellite. (c) State this relationship in algebraic form. 5 Calculate the orbital velocity required by the Space Shuttle when at an orbital altitude of 250 km above the surface of the Earth. Assume the mass of the Earth is kg and its radius is 6380 km. 6 Apollo command modules orbited the Moon at an altitude of 110 km. Calculate the orbital velocity required to do this. Assume that the mass of the Moon is kg and its radius is 1738 km. 7 Determine the orbital velocity of a GPS satellite orbiting at an altitude of km above the Earth. 8 A 5200 kg satellite orbits the Earth at an altitude of km. Its velocity is 4800 ms 1. Determine the gravitational force required to keep this satellite in orbit. 3.5 Gravitational fields Graphing gravitational force The gravitational force is an attractive force, whereas the force between electric charges can be either attractive or repulsive. For the force the Earth exerts on the Moon, there is a distance vector from the centre of the Earth to the centre of the Moon, whereas the force vector points in the opposite direction, back to the Earth. For this reason, the gravitational force equation has a negative sign and the force is therefore graphed under the distance axis. FIGURE 3.9 The Earth exerts gravitational force on the Moon. Earth Distance vector Force vector Moon 74 Jacaranda Physics 12

17 The straight blue line in the graph shows how the gravitational force by the Earth on you would decrease if you were to drill down to the centre of the Earth. Newton calculated that if you were inside a hollow sphere, the gravitational force from the mass in the shell would cancel out, no matter where you were inside the sphere. This means that if you were inside the Earth, only the mass in the inner sphere between you and the centre of the Earth would exert a gravitational force on you. This force will get smaller the closer to the centre you go, and at the centre of the Earth the gravitational force will be zero Drawing the gravitational field FIGURE 3.10 This diagram shows how the Earth s gravitational force varies with distance r from the Earth s centre. Earth Newton s Law of Universal Gravitation describes the force between two masses. However, the solar system GMm F g = R 2 has many masses, each attracting each other. The Sun, E g the heaviest object in the solar system, determines the orbits of all the other masses, but each planet can cause minor variations in the orbital paths of the other planets. Precise calculation of the path of a planet or comet becomes a complicated exercise with many gravitational forces needing to be considered. Physicists after Newton realised it was easier to determine for each point in space the total force that would be experienced by a unit mass, that is, 1 kilogram, at that point. This idea slowly developed and in 1849 Michael Faraday, in explaining the interactions between electric charges and between magnets, formalised the concept, calling it a field. FIGURE 3.11 Diagram of the Earth s gravitational field. 0 R E lines of equal field strength r TOPIC 3 Motion and gravitational Fields 75

18 A field is more precisely defined as a physical quantity that has a value at each point in space. For example, a weather map showing the pressure across Australia could be described as a diagram of a pressure field. This is an example of a scalar field. In contrast, gravitational, electric and magnetic fields are vector fields; they give a value to the strength of the field at each point in space, and also a direction for that field at that point. For example, the arrows in the diagram of the Earth s gravitational field show the direction of the field, and the density of the lines (how close together the lines are) indicates the strength of the field. On a small scale, such as the interior of a room, the field lines or lines of force appear parallel and point down since this is the direction FIGURE 3.12 The gravitational field within a room on of the force that would be experienced by a mass placed within the field. Earth. Of course, any large object near the Earth, such as the Moon, will have (a) In a room a gravitational field of its own, and the two fields will combine to form a more complex field, such as that shown in figure Note that there is g a point between the two, but somewhat closer to the Moon, at which the strength of the field is zero. In other words, the gravitational attraction of the Earth and that of the Moon are precisely equal but opposite in direction. Such points exist between any two masses but become noticeable when considering planets and stars that are close enough to be gravitationally bound together. FIGURE 3.13 The gravitational field around the Earth and Moon. The overall shape depends upon the relative strengths of the two fields involved. Earth Calculating gravitational field strength A value for the strength of the gravitational field around a mass M can be determined from the value of the force on a unit mass in the field. If the mass m in Newton s Universal Law of Gravitation equation is assigned a value of 1 kg, then the force expression will give the strength of the gravitational field: g = GM r 2 The unit of gravitational field strength is Newtons per kilogram, N kg 1. The strength of the gravitational field at the Earth s surface can be calculated using the values for the mass and radius of the Earth from table 3.2: g = = N m 2 kg 2 ( kg) ( m) 2 = 9.80 N kg 1 Moon 76 Jacaranda Physics 12

19 This is the acceleration due to gravity at the Earth s surface. FIGURE 3.14 Graph of the magnitude of the strength of the Earth s gravitational field. The value of the Earth s radius used here, 6380 km (at the equator), is an average value, so the value of g calculated, N kg 1, also represents an average value. In fact, due to the spin of the Earth, as well as the Earth being slightly flattened at the poles, g varies from a minimum value at the equator of N kg 1 to a maximum value 9.8 of N kg 1 at the poles. In addition, local variations 4 in g can occur due to variations in density and thickness of the Earth s crust. The formula for gravitational field strength indicates R E 2R E 3R E r (m) that the value of g will also vary with altitude above the Earth s surface. By using a value of r equal to the radius of the Earth (R E ) plus altitude, the values of g at different altitudes as shown in table 3.5 can easily be calculated using a modified version of the formula as follows: G M g = E (R E + altitude) 2 TABLE 3.5 The variation of g with altitude above Earth s surface. Altitude (km) g (N kg 1 ) Comment Earth s surface Mount Everest Arbitrary beginning of space Mercury capsule orbit altitude Typical space shuttle orbit altitude Communications satellite orbit altitude It is clear from table 3.5 that the effect of the Earth s gravitational field is felt quite some distance out into space. 3.5 SAMPLE PROBLEM 1 Determine the gravitational field strength experienced by an astronaut if she were (a) on the surface of Venus (The mass of Venus is kg and its radius is 6052 km) (b) in orbit around Saturn s largest moon, Titan, at an altitude of 150 km (Titan s mass is kg and its radius is 2575 km). SOLUTION: (a) g = G M Venus R Venus = N m 2 kg 2 ( kg) ( m) 2 = 8.87 N kg 1 The gravitational field strength would be 8.87 N kg 1 directed downwards towards the centre of Venus. Gravitational field strength (N kg 1 ) TOPIC 3 Motion and gravitational Fields 77

20 G M (b) g = Titan (R Titan + altitude) = N m 2 kg 2 ( kg) ( m m) 2 = 1.21 N kg 1 The gravitational field strength would be 1.21 N kg 1 directed downwards towards the centre of Titan. At the time Newton developed his Law of Universal Gravitation, he knew it did not provide an explanation for how gravity works, that is, how action at a distance was achieved. It is inconceivable... that Gravity should be innate, inherent and essential to Matter, so that one body may act upon another at a distance thro a Vacuum, without the Mediation of any thing else... is to me so great an Absurdity that I believe no Man who has in philosophical Matters a competent Faculty of thinking can ever fall into it. Gravity must be caused by an Agent acting constantly according to certain laws; but whether this Agent be material or immaterial, I have left to the Consideration of my readers. Newton, 1692 The concept of a field now provides an explanation for action at a distance. 3.5 Exercise 1 1 Draw the gravitational field resulting when two spheres of equal mass are placed near each other. 2 How would the gravitational field diagram for the Earth differ from that of Mars? 3 What is the value of the Earth s gravitational field strength: (a) in the centre of the Earth (b) at a distance R E from the Earth s centre (Hint: refer to figure 3.10) 2 (c) at an altitude of 4000 m above the Earth s surface? 4 What would be the effect on the gravitational field strength experienced on the Earth s surface if the Earth s radius were tripled but the mass remained constant? 5 A 2 kg object is dropped from a height of 10 m on the surface of the Earth. How much longer would the same object take to fall if dropped from the same height on the surface of the Moon? Assume the mass of the Moon = kg and the radius of the Moon = 1740 km. 3.6 Energy in a gravitational field Potential energy in a gravitational field Gravitational potential energy, U, is the energy of a mass due to its position within a gravitational field. Here on Earth, the U of an object at a height h above the ground is easily found as it is equal to the work done to move the object from the ground: U = work done to move to the point = force required distance moved = (mg) h = mg h Hence, in this case U = mgh. We chose the ground as our starting point because this is our defined zero level, that is, the place where U = 0. Note that since work must be done on the object to lift it, it acquires energy. Hence, at height h, U is greater than zero. 78 Jacaranda Physics 12

21 FIGURE 3.15 Different levels of U. If we choose a planet s surface as the zero level U x has a positive value. If infinity is chosen as the zero level, U has a negative value. Planet Work must be done to move a mass against a gravitational field. U at surface Point x < U at x < U at (= 0 by our definition) On a larger, planetary scale we need to rethink our approach. Due to the inverse square relationship in the Law of Universal Gravitation, the force of attraction between a planet and an object will drop to zero only at an infinite distance from the planet. For this reason, we will now choose infinity (or some point a very large distance away) as our level of zero potential energy. There is a strange side effect of our choice of zero level. Because gravitation is a force of attraction, work must be done on the object to move it FIGURE 3.16 A graph showing how the negative value for gravitational potential energy, U, increases with distance up to a maximum value of zero. from a point, X, to infinity; that is, against the field so that it gains potential energy. Therefore, U (potential energy at infinity) > U x (potential energy at point X) As U = 0, this means that U 0 > U x That is, U x has a negative value! Using the same approach as earlier, the gravitational potential energy U of an object at a point X, in a gravitational field is equal to the work done O r p r to move the object from the zero energy level at G Mm infinity (or some point very far away) to point X. r p It can be shown mathematically that: TOPIC 3 Motion and gravitational Fields 79

22 U = G M m r where M is the mass of the planet, m is the mass of the object and r is the distance separating the masses. 3.6 SAMPLE PROBLEM 1 Given the following data, determine the gravitational potential energy of: (a) the Moon within the Earth s gravitational field (b) the Earth within the Sun s gravitational field. Mass of the Earth = kg Mass of the Moon = kg Mass of the Sun = kg Earth Moon distance = m Earth Sun distance = m SOLUTION: (a) U = G M E m M r = N m 2 kg 2 ( kg) ( kg) ( m) = J That is, the gravitational potential energy of the Moon is approximately J. Put another way, the work that would be done in moving the Moon from a very large distance away from Earth to its current distance would be J. The negative sign indicates that this would be work done by the system (not on the system) in moving the Moon from a very large distance away from the Earth to its present orbital distance. This negative work represents potential energy lost by the system as the Moon and the Earth are brought together (converted into other forms of energy, most probably kinetic). Since the U is reduced below the zero level (see figure 3.16), it is quite appropriate that it should appear as a negative value. (b) U = G M S m E r = N m 2 kg 2 ( kg) ( kg) ( m) = J That is, the gravitational potential energy of the Earth is approximately J Kinetic energy of an orbiting object As we have seen in Section 3.4.1, the orbital velocity v o of an object as it moves around a mass M at a distance r from the mass s centre can be described by the equation: GM v 0 = r 80 Jacaranda Physics 12

23 This means that we can develop an equation describing the kinetic energy of an orbiting object as follows: E k = 1 2 mv2 0 = m G M ( r ) (where m is the mass of the orbiting object) = 1 2 m G M r So, E k = G M m 2 r Note that the object s kinetic energy is half that of its potential energy but opposite in sign. 3.6 SAMPLE PROBLEM 2 Calculate the kinetic energy of a 1500 kg satellite orbiting the Earth at a distance of km from the Earth s centre. Assume the mass of the Earth = kg. SOLUTION: E k = G M E m S 2r = N m 2 kg 2 ( kg) ( kg) 2 ( m) = J The satellite has a kinetic energy of Joules (or 9.95 GJ) Orbital energy Any satellite travelling in a stable circular orbit at a given orbital radius has a characteristic total mechanical energy E. This is the sum of its kinetic energy E k (due to its orbital velocity) and its gravitational potential energy U (due to its height). We can derive an expression for the total orbital energy E: E = U + E k = G M m + G M m r 2r = G M m 2r This equation looks very similar to the equation for Ep and also represents a negative energy well. The value of the mechanical energy of a satellite orbiting a planet depends only on the masses involved and the radius of the orbit. A lower orbit produces a more negative value of E and, therefore, less energy, while a higher orbit corresponds to more energy Escape velocity Isaac Newton wrote that it should be possible to launch a projectile fast enough so that it achieves an orbit around the Earth. His reasoning was that a stone thrown from a tall tower will cover a considerable range before striking the ground. If it is thrown faster, it will travel further before stopping. If thrown faster still, it TOPIC 3 Motion and gravitational Fields 81

24 will have an even greater range. If thrown fast enough then, as the stone falls, the Earth s surface curves away, so that the falling stone never actually lands on the ground, and orbits the Earth. It was only a thought experiment, of course. He had no way of testing this idea but it does hit upon one important fact that for any given altitude, there FIGURE 3.17 Newton s suggestion for is a specific velocity required for any object to achieve a stable circular orbit. achieving an orbit. If this specific velocity is exceeded slightly, then the object will follow an Tower E elliptical orbit around the Earth. If the specific velocity is exceeded further still, then the object will follow a parabolic or hyperbolic path away from the Earth. This is the manner in which space probes depart the Earth and head off into space. We will now consider a situation similar to Newton s. Imagine throwing a A Earth B stone directly up. When thrown, the stone will rise to a certain height before C falling back to Earth. If thrown faster, it will rise higher. If thrown fast enough, it should rise up and continue to rise, slowing down but never falling back to D Earth, and finally coming to rest only when it has completely escaped the Earth s gravitational field. The initial velocity required to achieve this is known as escape velocity. By considering the kinetic and gravitational potential energy of a projectile, it can be shown mathematically that the escape velocity of a planet depends only upon the universal gravitation constant and the mass and the radius of the planet. Escape velocity (v esc ) from a large body of mass M is achieved when the total orbital energy of an object is equal to 0, that is: E k + U = 0 This means that, for an object of mass m, 1 2 mv2 esc G M m = 0 r Cancelling for m and rearranging, we see v 2 esc = 2G M r 2G M v esc = r 3.6 SAMPLE PROBLEM 3 Calculate the escape velocity for Earth (assume mass of the Earth = kg and radius of the Earth = m). SOLUTION: v esc = 2G M E R E = N m 2 kg 2 ( kg) ( m) = m s km h 1 82 Jacaranda Physics 12